In the given circuit, R is a pure resistor, L is a pure inductor, S is a 100V, 50Hz AC source and A is an AC ammeter. With either K1 or K2 alone closed, the ammeter reading is I. If the source is changed to 100V, 100Hz, the ammeter reading with K1 alone closed and with K2 alone closed will be respectively

Engineering

Physics

AC Circuit Analysis

LCR and Resonance in AC Circuit

Question

In the given circuit, R is a pure resistor, L is a pure inductor, S is a 100V, 50Hz AC source and A is an AC ammeter. With either K₁ or K₂ alone closed, the ammeter reading is I. If the source is changed to 100V, 100Hz, the ammeter reading with K₁ alone closed and with K₂ alone closed will be respectively

I, $\frac{I}{2}$

I, 2I

2I, I

2I, $\frac{I}{2}$

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$I = \frac{100}{R} = \frac{100}{2 π (50) (L)}$

now at 100Hz
new I when K₁ is close only
$I_{1} = \frac{100}{R} = I$
and new I when only K₂ is closed
$I_{2} = \frac{100}{2 π \times (100) \times L} = \frac{I}{2}$